Categories: Physics, Plasma physics.

# Lawson criterion

For sustained nuclear fusion to be possible, the Lawson criterion must be met, from which some required properties of the plasma and the reactor chamber can be deduced.

Suppose that a reactor generates a given power $$P_\mathrm{fus}$$ by nuclear fusion, but that it leaks energy at a rate $$P_\mathrm{loss}$$ in an unusable way. If an auxiliary input power $$P_\mathrm{aux}$$ sustains the fusion reaction, then the following inequality must be satisfied in order to have harvestable energy:

\begin{aligned} P_\mathrm{loss} \le P_\mathrm{fus} + P_\mathrm{aux} \end{aligned}

We can rewrite $$P_\mathrm{aux}$$ using the definition of the energy gain factor $$Q$$, which is the ratio of the output and input powers of the fusion reaction:

\begin{aligned} Q \equiv \frac{P_\mathrm{fus}}{P_\mathrm{aux}} \quad \implies \quad P_\mathrm{aux} = \frac{P_\mathrm{fus}}{Q} \end{aligned}

Returning to the inequality, we can thus rearrange its right-hand side as follows:

\begin{aligned} P_\mathrm{loss} \le P_\mathrm{fus} + \frac{P_\mathrm{fus}}{Q} = P_\mathrm{fus} \Big( 1 + \frac{1}{Q} \Big) = P_\mathrm{fus} \Big( \frac{Q + 1}{Q} \Big) \end{aligned}

We assume that the plasma has equal species densities $$n_i = n_e$$, so its total density $$n = 2 n_i$$. Then $$P_\mathrm{fus}$$ is as follows, where $$f_{ii}$$ is the frequency with which a given ion collides with other ions, and $$E_\mathrm{fus}$$ is the energy released by a single fusion reaction:

\begin{aligned} P_\mathrm{fus} = f_{ii} n_i E_\mathrm{fus} = \big( n_i \expval{\sigma v} \big) n_i E_\mathrm{fus} = \frac{n^2}{4} \expval{\sigma v} E_\mathrm{fus} \end{aligned}

Where $$\expval{\sigma v}$$ is the mean product of the velocity $$v$$ and the collision cross-section $$\sigma$$.

Furthermore, assuming that both species have the same temperature $$T_i = T_e = T$$, the total energy density $$W$$ of the plasma is given by:

\begin{aligned} W = \frac{3}{2} k_B T_i n_i + \frac{3}{2} k_B T_e n_e = 3 k_B T n \end{aligned}

Where $$k_B$$ is Boltzmann’s constant. From this, we can define the confinement time $$\tau_E$$ as the characteristic lifetime of energy in the reactor, before leakage. Therefore:

\begin{aligned} \tau_E \equiv \frac{W}{P_\mathrm{loss}} \quad \implies \quad P_\mathrm{loss} = \frac{3 n k_B T}{\tau_E} \end{aligned}

Inserting these new expressions for $$P_\mathrm{fus}$$ and $$P_\mathrm{loss}$$ into the inequality, we arrive at:

\begin{aligned} \frac{3 n k_B T}{\tau_E} \le \frac{n^2}{4} \expval{\sigma v} E_\mathrm{fus} \Big( \frac{Q + 1}{Q} \Big) \end{aligned}

This can be rearranged to the form below, which is the original Lawson criterion:

\begin{aligned} n \tau_E \ge \frac{Q}{Q + 1} \frac{12 k_B T}{\expval{\sigma v} E_\mathrm{fus}} \end{aligned}

However, it turns out that the highest fusion power density is reached when $$T$$ is at the minimum of $$T^2 / \expval{\sigma v}$$. Therefore, we multiply by $$T$$ to get the Lawson triple product:

\begin{aligned} \boxed{ n T \tau_E \ge \frac{Q}{Q + 1} \frac{12 k_B T^2}{\expval{\sigma v} E_\mathrm{fus}} } \end{aligned}

For some reason, it is often assumed that the fusion is infinitely profitable $$Q \to \infty$$, in which case the criterion reduces to:

\begin{aligned} n T \tau_E \ge \frac{12 k_B T^2}{\expval{\sigma v} E_\mathrm{fus}} \end{aligned}

## References

1. M. Salewski, A.H. Nielsen, Plasma physics: lecture notes, 2021, unpublished.

© Marcus R.A. Newman, a.k.a. "Prefetch". Available under CC BY-SA 4.0.